SOME INEQUALITIES FOR THE TRIANGLE
INVOLVING FIBONACCI NUMBERS
JOSÉ LUIS DÍAZ-BARRERO
Abstract. In this note classical inequalities and Fibonacci numbers are used to obtain some miscellaneous inequalities involving
the elements of a triangle.
1. Introduction
The elements of a triangle are a source of many nice identities and
inequalities. A similar interpretation exists for Fibonacci numbers.
Many of these identities and inequalities have been documented in
extensive lists that appear in the work of Botema [1], Mitrinovic [3]
and Koshy [2]. However, as far as we know, miscellaneous geometric
inequalities for the elements of a triangle involving Fibonacci numbers
never have appeared. In this paper, using classical inequalities and
Fibonacci numbers some of these inequalities are given.
2. The Inequalities
In what follows some inequalities for the triangle are stated and proved.
We start with
Theorem 2.1. In all triangle 4ABC, with the usual notations, the
following inequality
p
(2.1) a2 Fn + b2 Fn+1 + c2 Fn+2 ≥ 4S Fn Fn+1 + Fn+1 Fn+2 + Fn+2 Fn
holds.
√
Proof. Let us denote by k = 4 Fn Fn+1 + Fn+1 Fn+2 + Fn+2 Fn . Then,
(2.1) reads
(2.2)
a2 Fn + b2 Fn+1 + c2 Fn+2 ≥ kS.
2000 Mathematics Subject Classification. 05A19,11B39.
Key words and phrases. Geometric inequalities, Miscellaneous triangle inequalities, Sequences of integers, Fibonacci numbers.
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JOSÉ LUIS DÍAZ-BARRERO
Taking into account the Cosine Law, we have
1
a2 Fn + b2 Fn+1 + (a2 + b2 − 2ab cos C)Fn+2 ≥ kab sin C,
2
or equivalently,
b
a
2 Fn + Fn+2 + 2 Fn+1 + Fn+2 − 4Fn+2 cos C + k sin C ≥ 0.
b
a
From Cauchy-Buniakovski-Schwarz’s inequality applied to (4Fn+2 , k)
and (cos C, sin C), we obtain
q
2
4Fn+2 cos C + k sin C ≤ 16Fn+2
+ k2.
On the other hand, from AM-GM inequality, we get
p
a
b
2 (Fn + Fn+2 ) + 2 (Fn+1 + Fn+2 ) ≥ 4 (Fn + Fn+2 )(Fn+1 + Fn+2 ).
b
a
Taking into account the preceding inequalities, we have
a
b
2 (Fn + Fn+2 ) + 2 (Fn+1 + Fn+2 ) − (4Fn+2 cos C + k sin C)
b
a
p
≥ 4 (Fn + Fn+2 )(Fn+1 + Fn+2 ) − (4Fn+2 cos C + k sin C)
q
p
2
≥ 4 (Fn + Fn+2 )(Fn+1 + Fn+2 ) − 16Fn+2
+ k2 ≥ 0
√
when k ≤ 4 Fn Fn+1 + Fn+1 Fn+2 + Fn+2 Fn .
Consequently, (2.1) holds and Theorem 1 is proved.
As an immediate consequence of the preceding result we obtain the
following inequality.
Corollary 2.2. In all triangle 4ABC, holds
(2.3)
a2 Fn + b2 Fn+1 + c2 Fn+2 ≥ 4S
n+2
X
!1/2
2
Fk2 − Fn+1
.
k=1
Proof. In fact, from Theorem 1 we have
p
a2 Fn + b2 Fn+1 + c2 Fn+2 ≥ 4S Fn Fn+1 + Fn+1 Fn+2 + Fn+2 Fn
q
2
= 4S Fn Fn+1 + Fn+2
q
2
= 4S F12 + F22 + . . . + Fn2 + Fn+2
.
INEQUALITIES FOR THE TRIANGLE
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Note that in the last expression we have used the fact that F12 + F22 +
. . . + Fn2 = Fn Fn+1 . Therefore,
!1/2
n+2
X
2
a2 Fn + b2 Fn+1 + c2 Fn+2 ≥ 4S
Fk2 − Fn+1
k=1
and the proof is complete.
Before stating our next result we give a Lemma that we will use further
on.
Lemma 2.3. Let x, y, z and a, b, c be strictly positive real numbers.
Then, holds
√
√
√ 2
3 yza2 + zxb2 + xyc2 ≥ a yz + b zx + c xy .
√ √ √ →
→
Proof. Let −
u =
yz, zx, xy and −
v = (a, b, c). By applying
Cauchy-Buniakovski-Schwarz’s inequality, we get
2
√ √ √ √ √ √ 2
yz, zx, xy · (a, b, c) ≤k
yz, zx, xy k k (a, b, c) k2
or equivalently,
√
√
√ 2
(2.4)
a yz + b zx + c xy ≤ (yz + zx + xy)(a2 + b2 + c2 ).
On the other hand, by applying the rearrangement inequality yields
a2 yz + b2 zx + c2 xy ≥ b2 yz + c2 zx + a2 xy,
a2 yz + b2 zx + c2 xy ≥ b2 xy + a2 zx + c2 yz.
Hence, the right hand side of (2.4) becomes
(yz + zx + xy)(a2 + b2 + c2 ) ≤ 3(yza2 + zxb2 + xyc2 )
and the proof is complete.
In particular, setting x = Fn , y = Fn+1 , and z = Fn+2 in the preceding
Lemma, we get the following
Theorem 2.4. If a, b and c are the sides of triangle 4ABC, then
3 Fn+1 Fn+2 a2 + Fn+2 Fn b2 + Fn Fn+1 c2
2
p
p
p
≥ a Fn+1 Fn+2 + b Fn+2 Fn + c Fn Fn+1 .
Finally, we will use the preceding result to state and prove the following
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JOSÉ LUIS DÍAZ-BARRERO
h π
, we
Theorem 2.5. Let 4ABC be a triangle, then for α ∈ 0,
2
have
p
p
p
Fn+1 Fn+2 cos(C − α) + Fn+2 Fn cos(B − α) + Fn Fn+1 cos(A − α)
π
≤ 2Fn+2 cos
−α .
3
Proof. By applying Botema inequality [1] and Theorem 2, we get
p
2
p
p
a Fn+1 Fn+2 + b Fn+2 Fn + c Fn Fn+1
≤ 3 Fn+1 Fn+2 a2 + Fn+2 Fn b2 + Fn Fn+1 c2 ≤ 3R2 (Fn +Fn+1 +Fn+2 )2 ,
and from it,
p
p
p
√
a Fn+1 Fn+2 + b Fn+2 Fn + c Fn Fn+1 ≤ R 3(Fn + Fn+1 + Fn+2 ).
Since a = 2R sin A, b = 2R sin B, and c = 2R sin C, then from the
preceding inequality, we obtain
(2.5)
p
p
p
√
Fn+1 Fn+2 sin A + Fn+2 Fn sin B + Fn Fn+1 sin C ≤ 3Fn+2 .
On the other hand, by applying the asymmetric trigonometric inequality of J. Wolstenholme ([3],[4]), we have
p
p
p
(2.6) Fn+1 Fn+2 cos A + Fn+2 Fn cos B + Fn Fn+1 cos C ≤ Fn+2 .
Multiplying (2.5) by tan α, adding it up to (2.6), and after simplification, yields
p
Fn+1 Fn+2 [cos A cos α + sin A sin α]
p
+ Fn+2 Fn [cos B cos α + sin B sin α]
p
+ Fn Fn+1 [cos C cos α + sin C sin α]
h
i
π
π
≤ 2Fn+2 cos cos α + sin sin α
3
3
and the proof is complete.
References
1. O. Botema et alt. Geometric Inequalities, Wolters Noordhoff Publishing,
Groningen, 1969.
2. T. Koshy. Fibonacci and Lucas Numbers with Applications, John Wiley &
Sons, Inc., New York, 2001.
3. D. S. Mitirinovic, J. E. Pecaric and V. Volenec. Recent Advances in Geometric
Inequalities, Kluwer Academic Publishers, Dordrecht, 1989.
4. J. Wolstenholme. Mathematical Problems, MacMillan and Co, London, 1878.
INEQUALITIES FOR THE TRIANGLE
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Applied Mathematics III, Universitat Politècnica de Catalunya,
Jordi Girona 1-3, C2, 08034 Barcelona. Spain
E-mail address: [email protected]